How do you divide # (8+2i)/(9-3i) # in trigonometric form?

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Answer 1 · 2018-06-24T12:22:52

#color(crimson)((8 + i2)/(9 - i3) = 0.8692(0.8411 - i 0.5408 )#

#z_1 = 8 + i 2, z_2 = 9 - i3#

#z_1 / z_2 =( r_1 / r_2) ( (cos theta_1 - theta_2) + i (sin theta_1 - theta_2))#

#r_1 = sqrt(8^2 + 2^2) = sqrt68#

#theta_1 = tan ^ -1 (2/8) = 14.31^@# in I quadrant

#r_2 = sqrt(9^2 + 3^2) = sqrt 90#

#theta_2 = tan ^ -1 (-3/9) = -18.43^@ = 341.57^@# in IV quadrant

#z_1 / z_2 = sqrt(68/90) (cos(341.57 - 14.31) + i sin(341.57 - 18.43))#

#z_1 / z_2 = 0.8692 (cos 327.26 + i sin 327.26)#

#color(crimson)((8 + i2)/(9 - i3) = 0.8692(0.8411 - i 0.5408 )#